Applied Optimization...Please Help!!
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Applied Optimization...Please Help!!

[From: ] [author: ] [Date: 12-11-01] [Hit: ]
0.35).Please help! I would really appreciate it!!-That book answer is simply incorrect!......
Find the point P on the parabola y=x^2 closest to the point (3,0).

The answer in the book says P is approximately (0.59, 0.35).

Please help! I would really appreciate it!!

-
That 'book answer' is simply incorrect!
The minimum distance from the parabola y = x^2 to the point (x,y) = (3,0) is d = sqrt(5) = 2.236. This point on the parabola is (x,y) = (1,1)
Proof:
Let d = distance from the parabola to the point (3,0)
Using the distance formula together with drawing the little right triangle showing altitude h, the distance from point (0,3) to the curve and the remaining distance along x from the altitude base to the point (3,0) compute
d^2 = x^4 + (3 - x)^2
d = ((x^4 + (3 - x)^2)^0.5
differentiate d wrt x to obtain a minimum:
dd/dx = (4x^3 + 2x - 6)/2(x^4 + x^2 - 6x + 9)^0.5
The numerator of this derivative = zero when x = +1, and the denominator is non zero.
From the distance formula, the minimum distance, d, = 5^0.5 = 2.236 which is SMALLER than the distance from the point (0.59,0.35). Obviously, something is either wrong or has been transcribed incorrectly.

Anyway, I hope this clarifies! Good luck!
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